x ³ = 1
x ³ - 1 = 0
x ³ - 1³ = 0
Factorize the right side as a difference of cubes:
(x - 1) (x ² + x + 1) = 0
Then
x - 1 = 0 or x ² + x + 1 = 0
The first equation yields x = 1 as a root.
For the other equation, rearrange and complete the square to get the other two roots,
x ² + x + 1/4 = -3/4
(x + 1/2)² = -3/4
x + 1/2 = ±√(-3/4) = ± i √3/2
x = -1/2 ± i √3/2
x = (-1 ± i √3)/2
Since complex roots occurs in conjugate pairs, taking the sum of these two roots eliminates the imaginary part:
(-1 + i √3)/2 + (-1 - i √3)/2 = (-1 - 1)/2 = -2/2 = -1
and adding to 1 indeed shows that the sum of roots is zero.